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In an industrial process $10\, kg$ of water per hour is to be heated from $20^o C$ to $80^o C$ . To do this steam at $200^o C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at $90^o C$. How many kg of steam is required per hour. (Specific heat of steam $= 0.5\, cal/g^o C$, Latent heat of vaporisation $= 540 cal/g)$
$1\, g$
$1 \,kg$
$10\, g$
$10\, kg$
Solution
Suppose $m$ $kg$ steam is required per hour
Heat is released by steam in following three steps
$(i)\,When\,{150^ \circ }C\,steam\,\xrightarrow[{{Q_1}}]{}{100^ \circ }C\,steam$
$\mathrm{Q}_{1}=\mathrm{mc}_{\text {steam }} \Delta \theta=\mathrm{m} \times 1(150-100)=50 \mathrm{m}$ cal
$(ii)\,When\,{150^ \circ }C\,steam\,\xrightarrow[{{Q_2}}]{}{100^ \circ }C\,water$
$\mathrm{Q}_{2}=\mathrm{mL}_{\mathrm{v}}=\mathrm{m} \times 540=540 \mathrm{m} \mathrm{cal}$
$(iii)\,When\,{100^ \circ }C\,water\xrightarrow[{{Q_3}}]{}{90^ \circ }C\,water$
$\mathrm{Q}_{3}=\mathrm{mc}_{\mathrm{w}} \Delta \theta=\mathrm{m} \times 1 \times(100-90)=10 \mathrm{m} \mathrm{cal}$
Hence total heat given by the steam
$\mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2}+\mathrm{Q}_{3}=600 \mathrm{mcal} \ldots .(i)$
Heat taken by $10 \mathrm{kg}$ water
$\mathrm{Q}^{\prime}=\mathrm{mc}_{\mathrm{w}} \Delta \theta=10 \times 10^{3} \times 1 \times(80-20)$
$=600 \times 10^{3} \mathrm{cal}$
Hence $Q=Q^{\prime} \Rightarrow 600 \mathrm{m}=600 \times 10^{3}$
$\Rightarrow \mathrm{m}=10^{3} \mathrm{gm}=1 \mathrm{kg}$
